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We begin by assuming cylindrical symmetry around the axis OO ’. Actually, this assumption is not precisely correct, for as part (b) of [link] shows, the view of the toroidal coil varies from point to point (for example, P 1 , P 2 , and P 3 ) on a circular path centered around OO ’. However, if the toroid is tightly wound, all points on the circle become essentially equivalent [part (c) of [link] ], and cylindrical symmetry is an accurate approximation.

With this symmetry, the magnetic field must be tangent to and constant in magnitude along any circular path centered on OO ’. This allows us to write for each of the paths D 1 , D 2 , and D 3 shown in part (d) of [link] ,

B · d l = B ( 2 π r ) .

Ampère’s law relates this integral to the net current passing through any surface bounded by the path of integration. For a path that is external to the toroid, either no current passes through the enclosing surface (path D 1 ), or the current passing through the surface in one direction is exactly balanced by the current passing through it in the opposite direction ( path D 3 ) . In either case, there is no net current passing through the surface, so

B ( 2 π r ) = 0

and

B = 0 (outside the toroid) .

The turns of a toroid form a helix, rather than circular loops. As a result, there is a small field external to the coil; however, the derivation above holds if the coils were circular.

For a circular path within the toroid (path D 2 ), the current in the wire cuts the surface N times, resulting in a net current NI through the surface. We now find with Ampère’s law,

B ( 2 π r ) = μ 0 N I

and

B = μ 0 N I 2 π r (within the toroid) .

The magnetic field is directed in the counterclockwise direction for the windings shown. When the current in the coils is reversed, the direction of the magnetic field also reverses.

The magnetic field inside a toroid is not uniform, as it varies inversely with the distance r from the axis OO ’. However, if the central radius R (the radius midway between the inner and outer radii of the toroid) is much larger than the cross-sectional diameter of the coils r , the variation is fairly small, and the magnitude of the magnetic field may be calculated by [link] where r = R .

Summary

  • The magnetic field strength inside a solenoid is
    B = μ 0 n I (inside a solenoid)

    where n is the number of loops per unit length of the solenoid. The field inside is very uniform in magnitude and direction.
  • The magnetic field strength inside a toroid is
    B = μ o N I 2 π r (within the toroid)

    where N is the number of windings. The field inside a toroid is not uniform and varies with the distance as 1/ r .

Conceptual questions

Is the magnetic field inside a toroid completely uniform? Almost uniform?

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Explain why B = 0 inside a long, hollow copper pipe that is carrying an electric current parallel to the axis. Is B = 0 outside the pipe?

If there is no current inside the loop, there is no magnetic field (see Ampère’s law). Outside the pipe, there may be an enclosed current through the copper pipe, so the magnetic field may not be zero outside the pipe.

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Problems

A solenoid is wound with 2000 turns per meter. When the current is 5.2 A, what is the magnetic field within the solenoid?

B = 1.3 × 10 −2 T

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A solenoid has 12 turns per centimeter. What current will produce a magnetic field of 2.0 × 10 −2 T within the solenoid?

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If a current is 2.0 A, how many turns per centimeter must be wound on a solenoid in order to produce a magnetic field of 2.0 × 10 −3 T within it?

roughly eight turns per cm

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A solenoid is 40 cm long, has a diameter of 3.0 cm, and is wound with 500 turns. If the current through the windings is 4.0 A, what is the magnetic field at a point on the axis of the solenoid that is (a) at the center of the solenoid, (b) 10.0 cm from one end of the solenoid, and (c) 5.0 cm from one end of the solenoid? (d) Compare these answers with the infinite-solenoid case.

Figure A is a cross section of a solenoid that shows three windings. The distance from the center to the winding is 1.5 centimeters. The distance between the windings is 20 centimeters. The point is located at the center axis of the solenoid, opposite to the second winding. Figure B is a cross section of a solenoid that shows three windings. The distance from the center to the winding is 1.5 centimeters. The distance between the windings is 20 centimeters. The point is located at the center axis of the solenoid, between the first and the second winding. Figure C is a cross section of a solenoid that shows three windings. The distance from the center to the winding is 1.5 centimeters. The distance between the windings is 20 centimeters. The point is located at the center axis of the solenoid, five centimeters below the first winding.
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Determine the magnetic field on the central axis at the opening of a semi-infinite solenoid. (That is, take the opening to be at x = 0 and the other end to be at
x = . )

B = 1 2 μ 0 n I

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By how much is the approximation B = μ 0 n I in error at the center of a solenoid that is 15.0 cm long, has a diameter of 4.0 cm, is wrapped with n turns per meter, and carries a current I ?

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A solenoid with 25 turns per centimeter carries a current I . An electron moves within the solenoid in a circle that has a radius of 2.0 cm and is perpendicular to the axis of the solenoid. If the speed of the electron is 2.0 × 10 5 m/s , what is I ?

0.0181 A

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A toroid has 250 turns of wire and carries a current of 20 A. Its inner and outer radii are 8.0 and 9.0 cm. What are the values of its magnetic field at r = 8.1 , 8.5 , and 8.9 cm?

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A toroid with a square cross section 3.0 cm × 3.0 cm has an inner radius of 25.0 cm. It is wound with 500 turns of wire, and it carries a current of 2.0 A. What is the strength of the magnetic field at the center of the square cross section?

0.0008 T

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Practice Key Terms 2

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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